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GRE (mathematics subject) question - 14% success

  1. Feb 14, 2014 #1
    Hey guys. This question is really bugging me. (Please see question #63 here: http://www.ets.org/s/gre/pdf/practice_book_math.pdf) - written below for your convenience.

    [itex]f(x) = xe^{-x^{2}-x^{-2}}, x \neq 0[/itex]
    [itex]f(x) = 0, x = 0 [/itex]

    (apologies for not knowing the itex command do write this as a single f(x))

    The question is, at how many values of [itex]x[/itex] does the graph of f(x) have a horizontal tangent line?

    My answer was "two" and the correct answer is "three".

    For those of you that like a challenge, only 14% of the testers answered this correctly and is the least correctly answered question of the whole test.

    Differentiating f(x) at all x != 0 would yield two real results (and two imaginary results - which is ignored). The real question IMO is what happens at x == 0.

    As [itex]x[/itex] approaches 0+, f(x) tends to 0+ and as [itex]x[/itex] approaches 0-, f(x) tends to 0-(since the exponential tends to 0 either way).

    So, how can there be a horizontal tangent line at 0? Thanks!
     
    Last edited: Feb 14, 2014
  2. jcsd
  3. Feb 14, 2014 #2

    mathman

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    Calculate f'(0). Since f'(0) = 0, tangent is horizontal.
     
  4. Feb 14, 2014 #3
    But, it's 0+ on one side and 0- on the other, and obviously "0" at 0. From a purist perspective (whatever that means), the tangent will never be "horizontal" - even though it can be made infinitesimally close to it. But, the question is not asking for the latter, but the former ("horizontal"). IMO, the tangent will always have a positive slop (to go from 0- to 0+), even if the slope can be made as arbitrarily small as needed.. No?
     
  5. Feb 14, 2014 #4

    pasmith

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    Exactly the same can be said of [itex]x^3[/itex] at zero. It's the difference between a local extremum and a point of inflection.
     
  6. Feb 14, 2014 #5

    AlephZero

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    Just differentiate the function at x = 0, using the limit definition of a derivative.

    You get zero, from whatever "perspective" you mean.

    This is no different from differentiating ##x^3## at x = 0.
     
  7. Feb 14, 2014 #6

    PeroK

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    What about the function $$f(x) = x^3$$
    Does that have a horizontal tangent at x = 0?
     
  8. Feb 14, 2014 #7

    Ray Vickson

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    No: ##\lim_{x \to 0+} f(x) = 0 = \lim_{x \to 0-} f(x)##. The notation ##0\pm## refers to a direction of approach, not to a final answer. If you think that ##0+ \neq 0-##, please tell us what you think the difference is.
     
    Last edited: Feb 14, 2014
  9. Feb 15, 2014 #8
    I get what you guys are saying. Ok, here's my process of convincing myself...

    As some of you have said [itex]h(x)=x^{3}[/itex] has similar properties as [itex]f(x)[/itex] (from the original post). I would have had the same reservations that I indicated for [itex]f(x)[/itex] with [itex]h(x)[/itex], except that the same would be present with [itex]g(x)=x^{2}[/itex] as well.

    What I mean that, even though [itex]g(x) = x^{2}[/itex] clearly has a min point, it's slope isn't really "0" anywhere - just that it can be arbitrarily made to 0, at [itex]x=0[/itex]. A slope by definition requires two points to calculate it. And going back to the basic definition of a limit we can see this. Same argument applies to both [itex]f(x)[/itex] and [itex]h(x)[/itex] as well. Even though they don't have a min/max, they can clearly have a slope that can be arbitrarily made close to 0 - and for our purposes can be said to be "horizontal", since that's the only definition of "horizontal" we can have for a "general curve".

    Alternatively, a "true" horizontal line would only be present for curves of the form y = C, where C is a constant or a step-wise function with such a y = C component where we wish to take the derivative. In these cases, the slope IS 0.. Not "can be arbitrarily made close to 0". But, when people speak of a "horizontal tangent line", the "interpretation" should be read as the latter and not the former.
     
    Last edited: Feb 15, 2014
  10. Feb 15, 2014 #9

    Ray Vickson

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    Where are you getting these misunderstandings? You seem to be making up your own version of mathematics that is different from what mathematicians have developed over the past 400 years. What you are doing is not calculus, but something else. Please jettison these ideas of yours now, before you take them out into the real world and do some damage.

    For both g(x) = x^2 and h(x) = x^3, the 'slope' is exactly zero at x = 0---no ifs, and or buts. There is nothing approximate about it: it is exactly equal zero, not just 'close' to zero. BTW: a slope does not require two points to calculate it; it is a limit, obtained when the two points merge into a single point. Finite-difference approximations to the slope do, indeed, require two nearby points, but that is not what we mean when we talk about the slope.
     
  11. Feb 15, 2014 #10
    Lets keep the discussion academic shall we? I don't plan on doing any "damage" with my calculus.

    In any case, if you go to the basic definition of a derivative {f(x + h) - f(x)} / h, where h -> 0, that IS taking two points on the curve. I am not "making it up" as you say. Just need to go to the basic definition of a derivative. It's best not to lose sight of where the derivative comes from when we equate that to a slope.

    dg(x) / dx = 2x, didn't come from thin air, but from the basic definition of a derivative, which does consider two adjacent points on the curve - even if the two points can be made arbitrarily close to 0 (i.e. lim h -> 0). There is a reason why the definition of a derivative uses the limit notation and doesn't simply say "h = 0", but says "lim h->0".

    You said: "a slope does not require two points to calculate it; it is a limit, obtained when the two points merge into a single point"

    The latter part of the above statement is exactly what I said - but it does involve two points on the curve, even if they can be made arbitrarily close to each other. They are still two distinct points separated on the x-axis by "h", where h tends to 0.

    The first part of your statement "a slope does not require two points to calculate it" doesn't make sense. A slope by definition, literal, mathematical, or otherwise, should require two points to calculate it. The derivative definition in calculus notation is using this concept by making these two points "merge into a single point" as you say - but that's just a very glossy interpretation. Technically, they are still two separate points (I mean just LOOK at the mathematical definition of a derivative) - even if the difference is infinitesimally small and can be made arbitrarily close to 0.

    You said: "For both g(x) = x^2 and h(x) = x^3, the 'slope' is exactly zero at x = 0---no ifs, and or buts. There is nothing approximate about it"

    I would love to see your proof where you come up with dg(x)/dx = 2x, and evaluated to 0 at x = 0, WITHOUT using two separate points, f(x+h) and f(x), on the curve in your derivation. If you do that, hats off to you and I will learn something new. :)
     
    Last edited: Feb 15, 2014
  12. Feb 15, 2014 #11

    Ray Vickson

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    Go back and read what I wrote. I said that a derivative (= slope) is a LIMIT as two points merge into one. Of course you need to start with two points, but then you take a limit. Proving that g'(x) = 2x is a simple exercise from Calculus 101, and of course you need to start with two points---I never said otherwise. (Admittedly my sentence about not needing two points to calculate was badly worded, and should probably not have been written. However, the other things I said spell out what I really meant.)

    My guess is that you do not really understand clearly what is meant by the 'limit' concept, but at this point I am out of here.
     
    Last edited: Feb 15, 2014
  13. Feb 15, 2014 #12
    I understand perfectly well what is meant by a "limit" concept, as I described above.

    Now, just to re-iterate what YOU said: (your words below, not mine - copy/pasted)

    "a slope does not require two points to calculate it" (few posts above)
    "derivative (= slope)" (from above post)

    Combining the two would result in:
    [a "derivative" does not require two points to calculate it]

    Now, please tell me.. Who is not understanding what a derivative is... As you seem to love references to Calculus 101, my advice to you is, please go back to a Calculus 101 textbook and review the definition of a derivative using the limit concept/notation. Then if you are interested, we can talk. Using a "limit" (fancy word of the day?) doesn't negate the fact that there are two separate points.. Just b/c they can be made arbitrarily close to each other doesn't take away this basic fact.

    For the record, I am quite respectful and thankful for all those that come to one of my threads and answer my questions on any forum. But, when a poster such as you, who do not appear to have a good grasp of basic calculus, starts to throw around insinuations that others don't understand the subject, I find that a bit offensive.

    As I said... Your words above (in quotes) - not mine.
     
  14. Feb 15, 2014 #13

    SammyS

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    The misunderstanding appears to be yours.

    The derivative of f(x) is a limit .

    ##\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} ##​

    For h(x) = x3 ,

    ##\displaystyle h'(x)=\lim_{h\to 0}\frac{(x+h)^3-x^3}{h} ##

    ##\displaystyle \quad\quad=\lim_{h\to 0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h} ##

    ##\displaystyle \quad\quad=\lim_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h} ##

    ##\displaystyle \quad\quad=\lim_{h\to 0}3x^2+3xh+h^2 ##

    ##\displaystyle \quad\quad =0##

    You might be confusing the tangent line with the secant line. No secant line for h(x) = x3 is exactly horizontal.
     
  15. Feb 15, 2014 #14
    All I am asking you guys to do is to respectfully consider what the word "limit" means.. As in please look under the hood. SammyS, your derivation above is obviously correct, and what I am trying to say in this thread is staring right back at you. f(x + h) and f(x) on the first line are two separate points. Yes.. the notion of limit says that "h" can be arbitrarily made close to 0, but it's not 0. That's the whole POINT of a limit. If "h" was 0, the expression would result in 0/0 - which is obviously not what we want.

    This distinction is somewhat important to what I was asking in my very first question. Regardless of how much you magnify the function at x = 0, and you can magnify it indefinitely, f(x+h) and f(x) WILL be two different values, irrespective of how close to 0 you make h (as long it's not exactly 0). In contrast, for a truly horizontal line, f(x+h) - f(x) WILL be "0".

    Again, just to reiterate, all I am asking is to re-evaluate my question by looking at first principles, and what it "means" to have a "horizontal" tangent. So, don't hide the problem by throwing around the word "limit".. Look under the hood to see what a "limit" is.
     
  16. Feb 15, 2014 #15

    PeroK

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    Okay, so the tangent is not horizontal. So, what angle is it inclined to?
     
  17. Feb 15, 2014 #16
    The slope is = {f(x + h) - f(x)} / h, where h tends to 0, from a general sense.

    Since SammyS was nice enough to type out the derivative for [itex]f(x) = x^{3}[/itex], the answer at x = 0 is [itex]h^{2}, [/itex], where h can be arbitrarily made close to 0.

    The LIMIT when evaluated is obviously 0.. No arguments here. But, IMO, that's not what the question was asking - unless you are telling me to interpret a "horizontal tangent line" as a limit.. In my view, it will never BE a horizontal tangent line.. Even if it can be made arbitrarily close to it. Too nitty picky? Thx.
     
    Last edited: Feb 15, 2014
  18. Feb 15, 2014 #17

    PeroK

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    There's nothing special about horizontal tangents. You could say the same about any function at any point (unless the function is a straight line). E.g. f(x) = x^2 at the point (1, 1). The tangent line there is not y = 2x by the same argument.

    Or, back to "horizontal" tangents, if they are not horizontal are they +ve or -ve gradients? If the tangent is not horizontal it must be +ve or -ve. It can't be both.

    So, in effect tangent lines cannot be found? Because, clearly, if they cannot have a defined gradient, then they cannot exist.
     
  19. Feb 15, 2014 #18

    vela

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    The tangent line is, in fact, a limit.

    http://en.wikipedia.org/wiki/Tangent#Tangent_line_to_a_curve
     
  20. Feb 15, 2014 #19

    haruspex

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    So pick the point at x = 0.01. What is the tangent here? Well, as you approach x=0.01 from one side, the limit of the slope will be whatever... and as you approach 0.01 from the other side you'll get the same limiting slope. But by your argument you cannot say that the slope of the tangent at x = 0.01 is ever that value - just arbitrarily close to it. Hence, you don't accept the very concept of a tangent.
    Your argument is not nitty-picky - it's just wrong.
     
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